Question

a 50.0 ml solution of ca(oh)₂ with an unknown concentration was titrated with 0.340 m hno₃. to reach the endpoint, a total of 28.4 ml of hno₃ was required. given that 0.00966 mol of hno₃ are used in the titration, what quantity in moles of ca(oh)₂ had to be present in the initial reaction?

Explanation:

Step1: Write the balanced chemical equation

$2HNO_3 + Ca(OH)_2=Ca(NO_3)_2 + 2H_2O$

Step2: Determine mole - ratio

The mole - ratio of $HNO_3$ to $Ca(OH)_2$ is 2:1.

Step3: Calculate moles of $Ca(OH)_2$

Let $n_{Ca(OH)_2}$ be the moles of $Ca(OH)_2$ and $n_{HNO_3}$ be the moles of $HNO_3$. From the mole - ratio, $n_{Ca(OH)_2}=\frac{n_{HNO_3}}{2}$. Given $n_{HNO_3} = 0.00966$ mol, so $n_{Ca(OH)_2}=\frac{0.00966}{2}=0.00483$ mol.

Answer:

$0.00483$